Uniform convergence of Fourier series

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We study some of the basic properties of the Fourier series. In particular, we show that if the series of the Fourier coefficients of \(f\) converges absolutely, then Fourier series \(S[f]\) converges uniformly to \(f\) at the continuous points of \(f\).



Lemma (Condition for the uniform convergence of Fourier series)

Let \(f \in \mathcal{R}_{2\pi}^{1}\) with its Fourier coefficients satisfying the conditions

\[\begin{eqnarray*} \sum_{n=0}^{\infty}|a_n| &<& +\infty,\\ \sum_{n=1}^{\infty}|b_n| &<& +\infty. \end{eqnarray*}\]

Then, the Fourier series \(S[f]\) converges uniformly with respect to \(x\in \mathbb{R}\). In particular, \(S[f]\) is continuous.

Proof. Consider the following positive (non-negative) term series:

\[\frac{1}{2}a_0 + \sum_{n=1}^{\infty}(|a_n| + |b_n|).\]

By assumption, this series converges. Furthermore, this is a dominating series of \(S[f]\). Thus, \(S[f]\) converges uniformly. All terms of \(S[f]\) are continuous (they are just \(\cos\) and \(\sin\)). Thus, by the Theorem on Uniform convergence and continuity, \(S[f]\) is continuous. ■

See also: Sequence of functions (Uniform convergence and continuity)

Lemma (Zero function)

If all the Fourier coefficients of \(f \in \mathcal{R}_{2\pi}^{1}\) are zero, that is,

\[\begin{cases} \int_{-\pi}^{\pi}f(x)\cos(nx)dx = 0 & (n = 0, 1, \cdots),\\ \int_{-\pi}^{\pi}f(x)\sin(nx)dx = 0 & (n = 1, 2, \cdots), \end{cases} \tag{eq:zero}\]

then \(f(x_0) = 0\) where \(x_0\) is any continuous point of \(f\). If, in addition to the condition (eq:zero), \(f(x)\) is a continuous function, then \(f(x) = 0\) (i.e., identically zero).

Proof. By changing variables \(x \mapsto x - x_0\), we may assume \(x_0 = 0\). Thus, we may assume that \(f(x)\) is continuous at \(x=0\), and we show that \(f(0) = 0\). We prove this by contradiction.

Suppose \(f(0) \neq 0\), in particular, \(f(0) > 0\). Then, Since \(f\) is continuous at \(x = 0\), there exists a \(\delta > 0\) such that, if \(|x| < \delta\), then \(|f(x) - f(0)| < \frac{f(0)}{2}\). Combining with the triangle inequality \(|f(0)| - |f(x)| \leq |f(x) - f(0)|\), we have

\[f(x) \geq \frac{1}{2}f(0).\]

Let us define the following auxiliary function

\[p(x) = 1 + \cos x - \cos \delta.\]

Assuming \(|x| \leq \pi\),

\[\begin{eqnarray} |x| \leq \delta &\implies& \cos x \geq \cos \delta \implies p(x) \geq 1,\\ |x| > \delta &\implies & - 1 \leq \cos x < \cos \delta \implies -\cos \delta \leq p(x) < 1\nonumber\\ &\implies& |p(x)| < 1. \tag{eq:px1} \end{eqnarray}\]

For \(k\in\mathbb{N}\), \(p(x)^k\) is a \(k\)-th degree polynomial in \(\cos x\). Meanwhile, \(\cos^k x\) can be expressed as a linear combination of \(1\), \(\cos x\), \(\cos(2x)\), \(\cdots, \cos(kx)\) (why?). Therefore, by (eq:zero),

\[\int_{-\pi}^{\pi}f(x)p(x)^kdx = 0 ~~~ (k = 1, 2, \cdots).\]

Splitting the interval of the integral on the left-hand side into \(|x|\leq \delta\) and \(|x| > \delta\),

\[0 = \int_{|x|\leq \delta}\cdot + \int_{|x| > \delta}\cdot\geq \int_{|x|\leq \delta}\frac{f(0)}{2}dx + \int_{|x| > \delta}f(x)p(x)^kdx.\]

Hence,

\[0 \geq f(0)\delta + \int_{|x| > \delta}f(x)p(x)^kdx.\tag{eq:0px}\]

We now calculate the limit of the above integral as \(k \to \infty\). If \(\delta < |x| < \pi\), by (eq:px1), we have

\[|f(x)p(x)^k| \leq |f(x)|,\]

and

\[\begin{eqnarray*} \int_{|x|>\delta}|f(x)|dx &< & +\infty,\\ \lim_{k\to\infty}f(x)p(x)^k &=& f(x)\cdot 0 = 0 ~~ (|x| > \delta). \end{eqnarray*}\]

By Arzelà's theorem,

\[\lim_{k\to\infty}\int_{|x|> \delta}f(x)p(x)^k dx = 0.\]

Applying this to (eq:0px), we have \(0 \geq f(0)\delta > 0\) which is a contradiction. Therefore, \(f(0)> 0\) is impossible. Similarly, we can show that \(f(0)< 0\) is impossible. Thus, \(f(0) = 0\). ■

See also: Arzelà's theorem

Theorem (If \(S[f]\) is uniformly convergent, then \(f=S[f]\))

If the Fourier series \(S[f]\) of \(f\in\mathcal{R}_{2\pi}^{1}\) converges uniformly, then \(S[f](x_0)\) converges to \(f(x_0)\) where \(x_0\) is any point at which \(f\) is continuous. In particular, if \(f\) is continuous with a period \(2\pi\) and \(S[f]\) converges uniformly, then \(f=S[f]\).

Proof. Let us define

\[g(x) = S[f](x) = \frac{1}{2}a_0 + \sum_{n=1}^{\infty}\{a_n\cos(nx) + b_n\sin(nx)\}.\]

Since the series on the right-hand side converges uniformly (by assumption), \(g(x)\) is continuous, and term-wise integration is allowed for the functions \(g(x)\cos(nx)\) and \(g(x)\sin(nx)\). Noting the orthogonality of sine and cosine functions, we have

\[\begin{eqnarray*} \int_{-\pi}^{\pi}g(x)\cos(nx)dx &=& \pi a_n ~~~ (n = 0, 1, \cdots),\\ \int_{-\pi}^{\pi}g(x)\sin(nx)dx &=& \pi b_n ~~~ (n = 1, 2, \cdots). \end{eqnarray*}\]

Combining these with the definition of Fourier coefficients of \(f\), we have

\[\begin{eqnarray*} \int_{-\pi}^{\pi}\{f(x) - g(x)\}\cos(nx)dx &=& 0~~~ (n = 0, 1, \cdots),\\ \int_{-\pi}^{\pi}\{f(x) - g(x)\}\sin(nx)dx &=& 0 ~~~ (n = 1, 2, \cdots). \end{eqnarray*}\]

Applying the above Lemma (zero function) to the function \(f - g\), we have \(f(x_0) - g(x_0) = 0\) where \(x_0\) is a continuous point of \(f\). ■

The next theorem should now be trivial.

Theorem

Let \(f\in\mathcal{R}_{2\pi}^{1}\). If the Fourier coefficients \(a_n\) and \(b_n\) of \(f\) satisfy \(\sum_{n}|a_n| < +\infty\) and \(\sum_{n}|b_n| < +\infty\), then the Fourier series \(S[f]\) of \(f\) converges uniformly and \(S[f](x) = f(x)\) at each continuous point \(x\) of \(f\).

Example. As seen above, the Fourier series in the above Example (evenfunc) converges uniformly. Therefore,

\[1 - \frac{|x|}{\pi} = \frac{1}{2} + \frac{4}{\pi^2}\left(\frac{\cos x}{1^2} + \frac{\cos(3x)}{3^2} + \frac{\cos(5x)}{5^2} + \cdots\right) ~~~ (-\pi < x < \pi).\]

Note that we now have "\(=\)" rather than "\(\sim\)." □




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